This is a very nice puzzle that kept me awake for a long time… ðŸ™‚

John and Dianne, two journalists (and experienced bridge players), are captured by some foreign government. The guardian says that he can free them, if they win the following game. First, Dianne must choose five random cards from a standard 52 card pack. Then, she must send one card at a time (through the guardian) to John. When John receives the fourth card, it must correctly guess the fifth card that Dianne still has. If he guesses correctly, they are both free. Otherwise, they will both stay in prison forever.

Can John and Dianne exit the prison in these conditions?

Assumptions:

– Before the actual game, John and Dianne can talk together for one day and decide their playing strategy. But, after that, they cannot communicate in any way – they are now completely isolated forever (or, at least, until they get out from the prison). Also, they are completely isolated from the outer world (no cell phones) etc. The only method to communicate between them will be through the cards that Dianne sends to John.

– The time interval between delivering the cards cannot be used as a method to communicate information. In fact, the guardian might randomly delay delivering the cards to John just to make sure that they are not using this trick to send additional information.

– Dianne cannot alter the cards the she sends in any way to send additional information. In fact, after receiving a card from Dianne, the guardian might send to John another (almost identical) card of the same type, to make sure that she didn’t use the card to send additional clues to John.

Hmm… Good puzzle.

With 5 cards, at least 2 will be the same suit. To reduce the choices, let’s send the LOWEST of that suit as the 5th card.

For the first card, lets send the SECOND lowest card of that same suit. So this reduces us to A-Q of a known suit, only 12 choices.

So we have 3 cards left to communicate the rank of the 5th card. The only variation that we have (assuming we can’t send them face-down vs face-up or something) is the sequence that we choose to send.

Unfortunately, 2^3 is only 8 choices, so that won’t quite work…

hmm… I’ll be back!

uh, make that 3

21 = only 6 orderings with only 3 cards! argg…I’d like a property P such that for every 5-element subset S of {0,1,…,51} there is a 4-element subset of S that is P and a 4-element subset of S that is not P. Then I could use the 2 choices of P and the 4!=24 orders of the given cards to select from 48 options. Since there are 52-4=48 unseen cards that’s exactly the right number.

You’re forgetting the bridge aspect. I’m wondering if that has something to do with the strategy.

To Nicolas Allen:

4 cards, when ordered, give 4!=24 possibilities.

The ‘trick’ is that she gets to CHOOSE which 4 of the 5 she sends, and which 1 is left.

This gives you the extra bit you need.

There are many ways of "selecting" which 1 to keep as the fifth card which will work, but it’s a rather academic question once you reach this point.

Mr Blobby- That is exactly what my post said. P is agreed upon before the game. By choosing which 4 cards to send you communicate either P or not P. That tells you whether the "extra bit" is a 0 (not P) or a 1 (P).

But I didn’t find it so easy that I could come up with many such P’s. Care to share a few?

Taking from BradC, we know atleast 2 cards will have the same suit. So we send one of those cards over first to indicate the suit. Then use the remaining 3 cards, we can communicate at most 6 orderings. The thing to notice is that given 2 cards of the same suit, they can at most be only 6 cards away from the other in either direction. So we must choose which of the same suit cards to send over wisely (i.e. the one that at most 6 away from the other). Good riddle ðŸ™‚

mah solution

I’ll explain it with an example

Pick the five cards –

Suppose they are 2H, 7S, 4D, 4C, JC

(H – Hearts, S – Spades, D – Diamonds, C – Clubs

J – Jack, Q – Queen, K – King, A – Ace)

First pick two cards with the same suit(there will always be at least two, there are four suits)

In this case they are 4C, JC. Separate them.

Now we are going to ‘value’ our cards in ascending order from 2,3….., Q,K,A

so Ace is highest

Now find smaller of the ‘distances’ between the two cards of the same suit

Counting forwards from 4C the distance to JC is 7

Counting backwards from 4C the distance to JC is 6 (3,2,A,K,Q,J)

Now the distance from JC to 4C counting forwards is also 6 (Note this)

We have 3 other cards which we can order in 6 different ways

(Values are 2..A, C,S,D,H Hearts highest 2 of any suit is always lower than 3 of any suit)

Now we decide that we will send the 3 permuted cards first, and one of the cards with the same suit fourth.

We will always send the fourth card such that we can reach the other card counting forwards.

So in this case the cards we send are

7S, 4D,2H, JC (first 3 are sixth possible permutation)

The prisoner who gets the cards finds the permutation is 6 and counts 6 from fourth card and reaches 4C

and thats it

Nerds are pansies. Why would you trust what the guardian tells you?

Grow a backbone and realise that a deck of cards can make a very effective weapon.

ok, Jock, thanks for your input… ?

So now that we’ve established that the two cards are at MOST 6 apart (allowing a wrap around from K-A-2), we just need to make sure we have a standard method for John and Dianne to number their permutations, so they get the same count.

First we need to order the cards.

Rohinton suggested 2..A, C,S,D,H

I’d probably choose A..K, D,C,H,S

but that’s just me.

Once we’ve ordered the cards, we will have a Low, Med, High. So all the permutations are:

1: L M H

2: L H M

3: M L H

4: M H L

5: H L M

6: H M L

Regarding the third assumption:

1) Would the card the guard sends be from the same 52-card deck?

Also, as for identification, is John required to guess the rank, the suit or both?

The previous puzzle was solved correctly in the comments. So I thought to present with another prison-style…

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